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I'm studying for a qualifying exam in algebra, and my abstract algebra skills are quite rusty. I'm attempting to solve the following problem:

Suppose that $\Phi:G\rightarrow\mathbb{Z}$ is a surjective group homomorphism from the abelian group $(G,+)$ to the group of integers under addition. Let $K$ be the kernel of $\Phi$, and let $g$ be an element of $G$ for which $\Phi(g)=1$. Prove that $G$ is the (internal) direct sum of $K$ and the subgroup of $G$ generated by $g$.

I was initially thrown off by the fact that $\Phi(g)=1$, thinking this meant $g\in K$ as $1$ is typically used to denote the identity. However, the codomain here is $(\mathbb{Z},+)$, so the identity is actually the additive identity of the integers, $0$. Thus $g\notin K$. Which I believe makes it clear that $K\cap\langle g\rangle=0$ by a simple homomorphism argument (something to the effect of $\Phi(ng)=\Phi(g)+\Phi(g)+\cdots+\Phi(g)=1+1+\cdots+1=n\neq0$). What I'm stuck on is how to argue that $G=K+\langle g\rangle$. I can use the first isomorphism theorem to claim that $\mathbb{Z}\cong G/K$ and $K\triangleleft G$, but I'm not sure how this helps.

Edit: I must be thinking of something wrong, as it seems to me that $\Phi(\langle g\rangle)=\mathbb{Z}_{\geq0}$. But this doesn't allow the desired result, as $\Phi$ is surjective, meaning everything in $\mathbb{Z}$ gets mapped to, so specifically, $\exists h\in G$ s.t. $\Phi(h)=-1$. There's no way to write $h=k+g'$ where $k\in K$ and $g'\in\langle g\rangle$.

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This is exactly the reason why I prefer to denote the identity of a general group by $e$. You can't confuse it with anything. In most books it is denoted by $e$ by the way.

Anyway, you want to show that $G=K+\langle g\rangle$. First of all it is clear that $K+\langle g\rangle\subseteq G$, because a sum of an element in $K$ and an element in $\langle g\rangle$ is a sum of two elements in $G$, hence it belongs to $G$. Now we want to show the other direction. Let $h\in G$. Denote $n=\Phi(h)$. Let's suppose $n>0$. Then:

$\Phi(h)=n=1+...+1=\Phi(g)+...+\Phi(g)=\Phi(ng)$.

Now denote $k=h-ng$. Then $\Phi(k)=\Phi(h)-\Phi(ng)=0$. So $h=k+ng$ where $k\in K,ng\in\langle g\rangle$.

Note that we assumed that $\Phi(h)>0$. Now if $\Phi(h)<0$ use the fact that $\Phi(-h)>0$, so $-h$ has the required representation. And if $\Phi(h)=0$ then $h\in K$ and then $h=h+0\in K+\langle g\rangle$.

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